A body is dropped from a certain height $H$ . If the ratio of the distances traveled by it in $(n - 3)$ seconds to $(n - 3) t h$ second is $4 : 3$ , find $H$ . $(T a k e g = 10 m s^{- 2})$

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Solution

$\frac{S_{n - 3}}{S_{n - 3} t h} = \frac{4}{3}$
$\frac{\frac{1}{2} g (n - 3)^{2}}{\frac{1}{2} g (2 (n - 3) - 1)} = \frac{4}{3}$
Solving above ,we get
$n = 5, \frac{11}{3}$ , but n=integer
$n = 5$
$H = \frac{1}{2} \times g \times 5^{2} = \frac{1}{2} \times 10 \times 25 = 125 m$

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A body is dropped from a certain height H. If the ratio of the distances traveled by it in (n−3) seconds to (n−3)th second is 4:3, find H. (Take g=10m s−2)

Sn−3Sn−3th=4312g(n−3)212g(2(n−3)−1)=43Solving above ,we getn=5,113, but n=integern=5H=12×g×52=12×10×25=125m

A body is dropped from a certain height $H$ . If the ratio of the distances traveled by it in $(n - 3)$ seconds to $(n - 3) t h$ second is $4 : 3$ , find $H$ . $(T a k e g = 10 m s^{- 2})$

$\frac{S_{n - 3}}{S_{n - 3} t h} = \frac{4}{3}$
$\frac{\frac{1}{2} g (n - 3)^{2}}{\frac{1}{2} g (2 (n - 3) - 1)} = \frac{4}{3}$
Solving above ,we get
$n = 5, \frac{11}{3}$ , but n=integer
$n = 5$
$H = \frac{1}{2} \times g \times 5^{2} = \frac{1}{2} \times 10 \times 25 = 125 m$